Suppose two groups on treatment and can be modelled by a Weibull distribution with a normal probability density function

\begin{align} f(x; \alpha, \lambda) = \alpha \lambda x^{\alpha - 1}e^{-\lambda x^{\alpha}}, \quad \text{where } x \geq \alpha \; \text{and } \lambda > 0 \end{align}

Where in the study $n_1$ participants are assigned to treatment 1, $x_i$, and $n_2$ participants are assigned to treatment 2, $y_j$. We assume $\alpha$ is known but not necessarily equal to 1. Group $X_i$ has a Weibull distribution with pdf $f(x_i; \alpha_1, \lambda_1)$ and group $y_j$ has a Weibull distribution with pdf $f(y_j; \alpha_2, \lambda_2)$.

Derive the likelihood equations and solve to find the MLEs of $\lambda_1$ and $\lambda_1$


Obtaining the joint likelihood distribution for both groups, $x_i$ and $y_j$, with sample $(x_1, x_2,...x_i, y_1, y_2,..., y_j)$ to obtain allows for the likelihood equation which depends on parameters $(\lambda_1, \lambda_2)$ to look like the following be as follows:

\begin{align*} L(\lambda_1, \lambda_2) &= L_{1}(\lambda_1) \cdot L_{2}(\lambda_2) \\ &= \prod_{i = 1}^{n}(\alpha_1\lambda_{1}x_i^{\alpha_i - 1}e^{-\lambda_1 x_i^{\alpha_1}}) \cdot \prod_{j = 1}^{n}(\alpha_2\lambda_{2}y_j^{\alpha_2 - 1}e^{-\lambda_1 y_j^{\alpha_2}}) \\ &= \alpha_1^{n_1}\lambda_1^{n_1}\prod_{i = 1}^{n} x_i^{\alpha_1 - 1}e^{-\lambda_1 \sum_{i = 1}^{n} x_i^{\alpha_1}} \cdot \alpha_2^{n_2}\lambda_2^{n_2}\prod_{j = 1}^{n}y_j^{\alpha_2 - 1} e^{-\lambda_2 \sum_{i = 1}^{n} y_j^{\alpha_2}} \end{align*}

This produces the following log-likelihood

\begin{align} \ell(\lambda_1, \lambda_2) = n_1\ln(\lambda_1) + n_2\ln(\lambda_2) - \lambda_1\sum_{i = 1}^{n} x_i^{\alpha_1} -\lambda_2\sum_{i = 1}^{n} y_j^{\alpha_2} \end{align}

The partial derivatives for $\lambda_1, \lambda_2$

For $\lambda_1$:

\begin{align} \frac{\partial{\ell(\lambda_1, \lambda_2)}}{\partial{\lambda_1}} = \frac{n_1}{\lambda_1} - \sum_{i = 1}^{n} x_i^{\alpha_1} \end{align}

For $\lambda_2$:

\begin{align} \frac{\partial{\ell(\lambda_1, \lambda_2)}}{\partial{\lambda_2}} = \frac{n_2}{\lambda_2} - \sum_{i = 1}^{n} y_j^{\alpha_2} \end{align}

My question.

From here, do I set the equations equal to 0 and solve independently for each to obtain $\hat{\lambda}_1, \hat{\lambda}_1$ or do I solve by setting both equal to 0 and solving in terms of each other.

For example independently: For $\lambda_1$:

\begin{align*} 0 &= \frac{\partial\ell(\lambda_1, \lambda_2)}{\partial\lambda_1} \\ 0 &= \frac{n_1}{\lambda_1} - \sum_{i = 1}^{n} x_i^{\alpha_1} \\ \hat{\lambda_1} &= \frac{n_1}{\sum_{i = 1}^{n} x_i^{\alpha_1}} \end{align*}

For $\lambda_2$: \begin{align*} 0 &= \frac{\partial\ell(\lambda_1, \lambda_2)}{\partial\lambda_1} \\ 0 &= \frac{n_2}{\lambda_2} - \sum_{i = 1}^{n} y_j^{\alpha_2} \\ \hat{\lambda_2} &= \frac{n_2}{\sum_{i = 1}^{n} y_j^{\alpha_2}} \end{align*}

Solving together:

For $\lambda_1$:

\begin{align} \hat{\lambda}_{1} = \frac{n_1\lambda_2}{\lambda_2\left(\sum_{i = 1}^{n} x_i^{\alpha_1} -\sum_{i = 1}^{n}y_j^{\alpha_2} \right) + n_2} \end{align}

For $\lambda_2$:

\begin{align} \hat{\lambda}_{2} = \frac{n_2\lambda_1}{\lambda_1\left(\sum_{i = 1}^{n} x_i^{\alpha_1} -\sum_{i = 1}^{n}y_j^{\alpha_2} \right) + n_1} \end{align}

Thank you for your help. This might be a trivial question but I would also like to understand why.


1 Answer 1


We want to solve for the MLEs in terms of the known parameters and the data, so you should solve independently for each $\hat\lambda$.


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